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Exercise - 4.1

Principle of Mathematical Induction

**Question-1 :-** Prove the following by using the principle of mathematical induction for all n ∈ N:

Let the given statement be P(n), i.e., P(n): 1 + 3 + 3^{2}+ ...... + 3^{n-1}= (3^{n}- 1)/2 For n = 1, we have, P(1): 1 = (3^{1}- 1)/2 = 1, which is true. Let P(k) be true for some positive integer k, i.e., P(n): 1 + 3 + 3^{2}+ ...... + 3^{k-1}= (3^{k}- 1)/2 We shall now prove that P(k + 1) is true. Now, 1 + 3 + 3^{2}+ ....... + 3^{k-1}+ 3^{k+1-1}= (1 + 3 + 3^{2}+.... + 3^{k-1}) + 3^{k}= (3^{k}- 1)/2 + 3^{k}= {(3^{k}- 1) + 2.3^{k}}/2 = {(1 + 2).3^{k}- 1}/2 = (3.3^{k}- 1)/2 = (3^{k+1}- 1)/2 Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

**Question-2 :-** Prove the following by using the principle of mathematical induction for all n ∈ N:

**Question-3 :-** Prove the following by using the principle of mathematical induction for all n ∈ N:

**Question-4 :-** Prove the following by using the principle of mathematical induction for all n ∈ N:

**Question-5 :-** Prove the following by using the principle of mathematical induction for all n ∈ N:

**Question-6 :-** Prove the following by using the principle of mathematical induction for all n ∈ N:

**Question-7 :-** Prove the following by using the principle of mathematical induction for all n ∈ N:

**Question-8 :-** Prove the following by using the principle of mathematical induction for all n ∈ N:

Let the given statement be P(n), i.e., P(n): 1.2 + 2.2^{2}+ 3.2^{2}+ ..... + n.2^{n}= (n – 1)^{2n+1}+ 2 For n = 1, we have, P(1): 1.2 = 2 = (1 – 1).2^{1+1}+ 2 = 0 + 2 = 2, which is true. Let P(k) be true for some positive integer k, i.e., 1.2 + 2.2^{2}+ 3.2^{2}+ ..... + k.2^{k}= (k – 1)^{2k+1}+ 2 .......(i) We shall now prove that P(k + 1) is true. Now, {1.2 + 2.2^{2}+ 3.2^{2}+ ..... + k.2^{k}} + (k+1)2^{k+1}= (k - 1).2^{k+1}+ 2 + (k + 1).2^{k+1}= 2^{k+1}{(k - 1) + (k + 1)} + 2 = 2^{k+1}. 2k + 2 = k.2^{(k+1)+1}+ 2 = {(k+1) - 1}.2^{(k+1)+1}+ 2 Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

**Question-9 :-** Prove the following by using the principle of mathematical induction for all n ∈ N:

Let the given statement be P(n), i.e., P(n): 1/2 + 1/4 + 1/8 + ......+ 1/2^{n}= 1 - 1/2^{n}For n = 1, we have P(1): 1/2 = 1 - 1/2^{1}= 1/2, which is true. Let P(k) be true for some positive integer k, i.e., 1/2 + 1/4 + 1/8 + ......+ 1/2^{k}= 1 - 1/2^{k}.......(i) We shall now prove that P(k + 1) is true. Now, (1/2 + 1/4 + 1/8 + ......+ 1/2^{k}) + 1/2^{k+1}= (1 - 1/2^{k}) + 1/2^{k+1}[using i] = 1 - 1/2^{k}+ 1/2.2^{k}= 1 - 1/2^{k}(1 - 1/2) = 1 - 1/2^{k}. 1/2 = 1 - 1/2^{k+1}Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

**Question-10 :-** Prove the following by using the principle of mathematical induction for all n ∈ N:

**Question-11 :-** Prove the following by using the principle of mathematical induction for all n ∈ N:

**Question-12 :-** Prove the following by using the principle of mathematical induction for all n ∈ N:

**Question-13 :-** Prove the following by using the principle of mathematical induction for all n ∈ N:

**Question-14 :-** Prove the following by using the principle of mathematical induction for all n ∈ N:

Let the given statement be P(n), i.e., P(n) : (1 + 1/1).(1 + 1/2).(1 + 1/3)......(1 + 1/n) = (n + 1) For n = 1, we have (1 + 1/1) = (1 + 1) = 2, which is true. Let P(k) be true for some positive integer k, i.e., P(k): (1 + 1/1).(1 + 1/2).(1 + 1/3) ......(1 + 1/k) = (k + 1) We shall now prove that P(k + 1) is true. Now, {(1 + 1/1).(1 + 1/2).(1 + 1/3)...... (1 + 1/k)} + {1 + 1/(k+1)} = (k + 1).{1 + 1/(k+1)} = (k + 1).{(k+1) + 1}/(k+1) = (k+1) + 1 Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

**Question-15 :-** Prove the following by using the principle of mathematical induction for all n ∈ N:

**Question-16 :-** Prove the following by using the principle of mathematical induction for all n ∈ N:

**Question-17 :-** Prove the following by using the principle of mathematical induction for all n ∈ N:

**Question-18 :-** Prove the following by using the principle of mathematical induction for all n ∈ N:

Let the given statement be P(n), i.e., P(n) : 1 + 2 + 3 + .......+ n < 1/8.(2n + 1)^{2}It can be noted that P(n) is true for n = 1 since 1 < 1/8.(2.1 + 1)^{2}= 9/8. Let P(k) be true for some positive integer k, i.e., 1 + 2 + 3 + .....+ k < 1/8.(2k + 1)^{2}......(1) We shall now prove that P(k + 1) is true whenever P(k) is true. Now, (1 + 2 + 3 + ... + k) + (k + 1) < 1/8.(2k + 1)^{2}+ (k + 1) [By using 1] < 1/8{(2k + 1)^{2}+ 8(k + 1)} < 1/8{(4k^{2}+ 1 + 4k + 8k + 8)} < 1/8{(4k^{2}+ 12k + 9)} < 1/8.(2k + 3)^{2}< 1/8{2(k+1) + 1}^{2}Hence, (1 + 2 + 3 + ... + k) + (k + 1) < 1/8.(2k + 1)^{2}+ (k + 1) Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

**Question-19 :-** Prove the following by using the principle of mathematical induction for all n ∈ N:

n(n + 1)(n + 5) is a multiple of 3.

Let the given statement be P(n), i.e., P(n): n (n + 1) (n + 5), which is a multiple of 3. It can be noted that P(n) is true for n = 1 since 1 (1 + 1) (1 + 5) = 12, which is a multiple of 3. Let P(k) be true for some positive integer k, i.e., k (k + 1) (k + 5) is a multiple of 3. Therefore, k (k + 1) (k + 5) = 3m, where m ∈ N ...... (1) We shall now prove that P(k + 1) is true whenever P(k) is true. Now, (k + 1){(k + 1) + 1}{(k + 1) + 5} = (k + 1)(k + 2){(k + 5) + 1} = (k + 1)(k + 2)(k + 5) + (k + 1)(k + 2) = {k(k + 1)(k + 5) + 2(k + 1)(k + 5) + (k + 1)(k + 2)} = 3m + (k + 1){2(k + 5) + (k + 2)} = 3m + (k + 1){2k + 10 + k + 2} = 3m + (k + 1){3k + 12} = 3m + 3(k + 1)(k + 4) = 3{m + (k + 1)(k + 4)} = 3q where q = m + (k + 1)(k + 4) is a natural number. Therefore, (k + 1){(k + 1) + 1}{(k + 1) + 5} is divisible by 3. Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

**Question-20 :-** Prove the following by using the principle of mathematical induction for all n ∈ N:

10^{2n-1} + 1 is divisible by 11.

Let the given statement be P(n), i.e., P(n): 10^{2n-1}+ 1 is divisible by 11. It can be observed that P(n) is true for n = 1 since P(1) = 10^{2.1-1}+ 1 = 11, which is divisible by 11. Let P(k) be true for some positive integer k, i.e., 10^{2k-1}+ 1 is divisible by 11. Therefore, 10^{2k-1}+ 1 = 11m, where m ∈ N ........ (1) We shall now prove that P(k + 1) is true whenever P(k) is true. Now, 10^{2(k+1)-1}+ 1 = 10^{2k+1}+ 1 = 10^{2k}. 10^{2}+ 1 = 10^{2}(10^{2k}+ 1 - 1) + 1 = 10^{2}(10^{2k}+ 1) - 10^{2}+ 1 = 10^{2}. 11m - 100 + 1 = 100 . 11m - 99 = 11(100m - 9) = 11r, where r = 100m - 9 is some natural number. Therefore, 10^{2(k+1)-1}+ 1 is divisible by 11. Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

**Question-21 :-** Prove the following by using the principle of mathematical induction for all n ∈ N:

x^{2n} – y^{2n} is divisible by x + y.

Let the given statement be P(n), i.e., P(n): x^{2n}– y^{2n}is divisible by x + y. It can be observed that P(n) is true for n = 1. This is so because x^{2 x 1}– y^{2 x 1}= x^{2}– y^{2}= (x + y) (x – y) is divisible by (x + y). Let P(k) be true for some positive integer k, i.e., x^{2k}– y^{2k}is divisible by x + y. Therefore, x^{2k}– y^{2k}= m (x + y), where m ∈ N ........ (1) We shall now prove that P(k + 1) is true whenever P(k) is true. Now, x^{2(k+1)}- y^{2(k+1)}= x^{2k}x^{2}- y^{2k}y^{2}= x^{2}(x^{2k}- y^{2k}+ y^{2k}) - y^{2k}y^{2}= x^{2}(m(x + y) + y^{2k}) - y^{2k}y^{2}= x^{2}. m(x + y) + y^{2k}. x^{2}- y^{2k}y^{2}= x^{2}. m(x + y) + y^{2k}. (x^{2}- y^{2}) = x^{2}. m(x + y) + y^{2k}. (x - y)(x + y) = (x + y)[x^{2}. m + y^{2k}. (x - y)] which is a factor of (x + y) Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

**Question-22 :-** Prove the following by using the principle of mathematical induction for all n ∈ N:

3^{2n+2} – 8n – 9 is divisible by 8.

Let the given statement be P(n), i.e., P(n): 3^{2n+2}– 8n – 9 is divisible by 8. It can be observed that P(n) is true for n = 1 since 32 × 1 + 2 – 8 × 1 – 9 = 64, which is divisible by 8. Let P(k) be true for some positive integer k, i.e., 3^{2k+2}– 8k – 9 is divisible by 8. Therefore, 3^{2k+2}– 8k – 9 = 8m; where m ∈ N ........(1) We shall now prove that P(k + 1) is true whenever P(k) is true. Now, 3^{2(k+1)+2}– 8(k+1) – 9 = 3^{2k+2}. 3^{2}– 8k - 8 – 9 = 3^{2}(3^{2k+2}+ 9 + 8k - 9 - 8k) – 8k - 17 = 3^{2}(3^{2k+2}- 9 - 8k) + 3^{2}(8k + 9) – 8k - 17 = 9.8m + 9(8k + 9) - 8k - 17 = 9.8m + 72k + 81 - 8k - 17 = 9.8m + 64k + 64 = 8(9m + 8k + 8) = 8r where r = 9m + 8k + 8 is a natural number Therefore, 3^{2(k+1)+2}– 8(k+1) – 9 is divisible by 8. Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

**Question-23 :-** Prove the following by using the principle of mathematical induction for all n ∈ N:

41^{n} – 14^{n} is a multiple of 27.

Let the given statement be P(n), i.e., P(n): 41^{n}– 14^{n}is a multiple of 27. It can be observed that P(n) is true for n = 1 since 41^{1}- 14^{1}= 27, which is a multiple of 27. Let P(k) be true for some positive integer k, i.e., 41^{k}– 14^{k}is a multiple of 27 Therefore, 41^{k}– 14^{k}= 27m, where m ∈ N ....... (1) We shall now prove that P(k + 1) is true whenever P(k) is true. Now, 41^{k+1}- 14^{k+1}= 41^{k}. 41 - 14^{k}. 14 = 41.(41^{k}+ 14^{k}- 14^{k}) - 14^{k}. 14 = 41.(41^{k}- 14^{k}) + 41 . 14^{k}- 14^{k}. 14 = 41.27m + 14^{k}(41 - 14) = 41.27m + 14^{k}. 27 = 27(41m + 14^{k}) = 27 x r where r = (41m + 14^{k}) is a natural number Therefore, 41^{k+1}– 14^{k+1}is a multiple of 27 Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

**Question-24 :-** Prove the following by using the principle of mathematical induction for all n ∈ N:

(2n + 7) < (n + 3)^{2}

Let the given statement be P(n), i.e., P(n): (2n + 7) < (n + 3)^{2}It can be observed that P(n) is true for n = 1 since 2.1 + 7 = 9 < (1 + 3)^{2}= 16, which is true. Let P(k) be true for some positive integer k, i.e., (2k + 7) < (k + 3)^{2}+ 2 .....(1) We shall now prove that P(k + 1) is true whenever P(k) is true. Now, 2(k + 1) + 7 = (2k + 7) + 2 Therefore, 2(k + 1) + 7 = (2k + 7) + 2 < (k + 3)^{2}2(k + 1) + 7 < k^{2}+ 6k + 9 + 2 2(k + 1) + 7 < k^{2}+ 6k + 11 Now, k^{2}+ 6k + 11 < k^{2}+ 8k + 16 Therefore, 2(k + 1) + 7 < (k + 4)^{2}2(k + 1) + 7 < {(k + 1) + 1}^{2}Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

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